In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30 km 25.0º south of west; then 5.10 km straight east; then 1.70 km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?
Solution:
Gilligan’s displacement is characterized by 7 vectors. To determine his final position relative to the starting point, we simply need to add the vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:
Vector | X-Component | Y-Component |
(1) | -2.5\:\cos 45^{\circ} =-1.7678\:\text{km} | +2.5\:\sin 45^{\circ} =+1.7678\:\text{km} |
(2) | +4.70\:\cos 60^{\circ} =+2.3500\:\text{km} | -4.70\:\sin 60^{\circ} =-4.0703\:\text{km} |
(3) | -1.30\:\cos 25^{\circ} =-1.1782\:\text{km} | -1.30\:\sin 25^{\circ} =-0.5494\:\text{km} |
(4) | +5.1000\:\text{km} | 0 |
(5) | +1.70\:\sin 5^{\circ} =+0.1482\:\text{km} | +1.70\:\cos 5^{\circ} =+1.6935\:\text{km} |
(6) | -7.20\:\cos 55^{\circ} =-4.1298\:\text{km} | -7.20\:\sin 55^{\circ} =-5.8979\:\text{km} |
(7) | +2.80\:\cos 10^{\circ} =+2.7575\:\text{km} | +2.80\:\sin 10^{\circ} =+0.4862\:\text{km} |
Sum | 3.2799\:\text{km} | -6.5701\:\text{km} |
The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. This is also the x and y components of the resultant vector.
To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:
\begin{align*} \text{R} & =\sqrt{\left(3.2799\:\text{km}\right)^2+\left(-6.5701\:\text{km}\right)^2} \\ \text{R} & =7.34\:\text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}
The direction of the resultant is calculated as follows:
\begin{align*} \theta & =\tan ^{-1}\left(\frac{6.5701}{3.2799}\right) \\ \theta & =63.47^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}
Therefore, Gilligan is about 7.34 km directed 63.47° South of East from the starting island.
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